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BigorU [14]
4 years ago
12

Katya is a ranger in a nature reserve in Siberia, Russia, where she studies the changes in the reserve's bear population size ov

er time. She found that the number of bears, B, in the reserve at t years since the beginning of the study can be modeled by the following function: B(t)=5000*2^(-0.05t) How long will it take the bear population in the reserve to get to 2000 bears? Round your answer, if necessary, to 2 decimal places.
Mathematics
2 answers:
miss Akunina [59]4 years ago
6 0
B(t) = 5000 * 2^(-0.05t) 

<span>2000 = 5000 * 2^(-0.05t) , div by 5000 </span>
<span>0.4 = 2^(-0.05t) , ln on both sides => </span>

<span>ln0.4 = (-0.05t)* ln2 </span>
<span>(ln0.4)/ln2) /(-0.05) = t </span>

<span>t = 26.43 </span>
<span>After 27 years</span>
olga_2 [115]4 years ago
6 0

Answer: 26.44 years since the begginig of the study.

Step-by-step explanation:

The populations of bears can be described by:

B(t) = 5000*2^(-0.05*t)

where t is in years.

we want to find t such:

5000*2^(-0.05*t) = 2000

2^(-0.05*t) = 2000/5000 = 2/5

then we use the rule:

a^y = x

then:

y = logₐ(x)

we use it in our equation and get:

2^(-0.05*t) = 2/5

-0.05*t = log₂(2/5) = ln(2/5)/ln(2)

t = ln(2/5)/ln(2) /-0.05 = 26.44

so 26.44 years since the begginig of the study.

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Answer:

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Step-by-step explanation:

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Answer:

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Soft-drink cans are filled by an automated filling machine and the standard deviation is 0.5 fluid ounces assume that the fill v
evablogger [386]
A.) For n independent variates with the same distribution, the standard deviation of their mean is the standard deviation of an individual divided by the square root of the sample size: i.e. s.d. (mean) = s.d. / sqrt(n)
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b.) In a normal distribution, P(X < x) is given by P(z < (x - mean) / s.d).
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Odd function:

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We need to find out the effects on the y-intercept when shifting the function  into:

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1.2. Effects on the regions where the graph is increasing and decreasing

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2. When  becomes  

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An example is shown in Figure 1. The graph in blue is the function:

and the function in red is:

So you can see that:

2.2. Effects on the regions where the graph is increasing and decreasing

The effects are the same just as in the previous case. So the new function increases and decreases in the same intervals of

In Figure 1 you can see that both functions increase at:

and decrease at:

2.3 The end behavior when the following changes are made.

It happens the same, the output is one unit less than the output of . So, you can write the points just as they were written before.

So you can realize this concept by taking a point with the same x-coordinate of both graphs in Figure 1.

FOR EVEN FUNCTIONS:

3. When  becomes  

3.1 Effects on the y-intercept

We need to find out the effects on the y-intercept when shifting the function  into:

As we know, the graph  intersects the y-axis when , therefore:

And:

So the new y-intercept is the negative of the previous intercept shifted one unit upward.

3.2. Effects on the regions where the graph is increasing and decreasing

In the intervals when the function  increases, the function  decreases. On the other hand, in the intervals when the function  decreases, the function  increases.

3.3 The end behavior when the following changes are made.

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FOR ODD FUNCTIONS:

4. When  becomes  

4.1 Effects on the y-intercept

In this case happens the same as in the previous case. The new y-intercept is the negative of the previous intercept shifted one unit upward.

4.2. Effects on the regions where the graph is increasing and decreasing

In this case it happens the same. So in the intervals when the function  increases, the function  decreases. On the other hand, in the intervals when the function  decreases, the function  increases.

4.3 The end behavior when the following changes are made.

Similarly, each point of the function  has the same x-coordinate just as the function  and the y-coordinate is the negative of the previous coordinate shifted one unit upward.

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